Chemical Kinetics

1. Chemical Kinetics

2. The Rate of Reaction • Chemical kinetics is concerned with the rate at which chemical reactions take place • The rate of reaction is the change in concentration of a reactant or a product over time (in M/s)

3. The Rate of Reaction • For the reaction A  B • We want the rate of reaction to be positive, so for a reactant, we put a negative sign since

4. The Rate of Reaction • For the reaction 2A  B : since A disappears twice as fast as B appears • In general, for the reaction aA + bB  cC + dD • Example: Write the rate expression for the following reaction: • Solution:

5. The Rate of Reaction • The rate of reaction can be determined from a graph of the concentration of a reactant or product versus time • The rate of reaction at a particular time is given by the slope of the tangent at this point • The rate of reaction decreases with time as the reactants are depleted

6. The Rate Law • The rate law relates the rate of reaction to the concentrations of reactants and a proportionality constant (the rate constant) • The effect of the concentration of a reactant is best determined by measuring the initial rate of reaction • The rate of the inverse reaction (products  reactants) is negligible as there are not yet any products to react • To determine the effect of the concentration of a reactant on the reaction rate, the concentrations of other reactants must be constant

7. The Rate Law • eg.; F2(g) + 2 ClO2(g)  2 FClO2(g) [F2] (M) [ClO2] (M) Initial Rate (M/s) 0.10 0.010 1.2 x 10-3 0.10 0.040 4.8 x 10-3 0.20 0.010 2.4 x 10-3 • If we keep [F2] constant, it is observed that the initial rate increases by a factor of four if [ClO2] increases by a factor of four • If we keep [ClO2] constant, it is observed that the initial rate increases by a factor of two if [F2] increases by a factor of two

8. The Rate Law • k is the rate constant for the reaction • You can take any of the points from the empirical data to find the value of k • Taking the first: • This method is called the isolation method

9. The Rate Law • In general, for the reaction • N.B. most of the time: • The sum of the exponents in the rate law (in this example, x + y) is the global order of the reaction • In this example, the reaction is of order x for A and of order y for B

10. The Rate Law • Example: We measured rates of the reaction A + 2 B  C at 25oC. Using this data, determine the rate law and the rate constant for this reaction. Experiment [A] initial [B] initial initial rate (M/s) 1 0.100 0.100 5.50 x 10-6 2 0.200 0.100 2.20 x 10-5 3 0.400 0.100 8.80 x 10-5 4 0.100 0.300 1.65 x 10-5 5 0.100 0.600 3.30 x 10-5

11. The Rate Law • Solution: In the second and third experiments, the initial [B] is constant, and it is seen that the rate quadruples if we double [A]: In the fourth and fifth experiment, the initial [A] is constant, and it is seen that the rate doubles if we double [B]: Thus, the rate law is: N.B. We could have used other pairs of experiments too. To determine the value of k, we use the first experiment:

12. Concentration as a Function of Time • Rate laws allow us to determine the concentrations of reactants at any moment in time • For example, for a reaction of global first order that is first order with respect to A: where [A]o is the concentration of A at t=0 (which need not necessarily the beginning of the experiment)

13. Concentration as a Function of Time • The formula is the integrated rate law for a first order reaction • We can modify the formula • This form of integrated rate law tells us that a graph of ln[A] versus time is a straight line with a slope of -k • Experimentally, we determine k this way

14. Concentration as a Function of Time • Example: The reaction 2A  B is first order with respect to A, and the rate constant is 2.8 x 10-2 s-1 at 80oC. How much time (in seconds) would it take for [A] to decrease from 0.88 M to 0.14 M?

15. Half-life • The half-life of a reaction, t1/2, is the time required for the concentration of a reactant to decrease by half • For a reaction of global first order: • For a reaction of global first order, the half-life is independent of the initial concentration

16. Half-life • Example: The half-life of a first order reaction is 84.1 min. Calculate the rate constant of the reaction. • Solution: or

17. Half-life • Example: 14C is a radioactive isotope that incorporates itself within living organisms. The half-life of 14C is 5730 years (The nuclear disintegration is a global first order reaction). If we find a piece of cloth produced from a previously living source that contains only 10% of the 14C that it would contain if it was living, what is the age of this piece of cloth?

18. The Effect of Temperature on the Rate Constant • When the temperature increases, reaction rates increase, and thus the rate constants increase • There are very few exceptions to this rule • One exception is a reaction catalyzed by an enzyme (at high temperatures, the enzyme is denatured and will no longer function properly)

19. Activation Energy • The activation energy is the energy barrier separating reactants from products (the reaction is either endothermic or exothermic) • When the temperature is increased, the fraction of molecules with a kinetic energy greater than the activation energy increases • Increasing the temperature increases the chance that a collision will have sufficient kinetic energy

20. The Arrhenius Equation • The Arrhenius equation expresses the rate constant as: where Ea is the activation energy, R=8.3145 JK-1mol-1, T is the temperature in Kelvin, and A is the frequency factor (the frequency of collisions) • The value of A does not vary significantly with varying temperatures • A graph of ln k versus 1/T has a slope of -Ea/R

21. The Arrhenius Equation • We can determine the value of the activation energy using the two rate constants, k1 and k2, at temperatures T1 and T2 • We subtract the first equation from the second

22. The Arrhenius Equation • Example: The rate constant of the (first order) reaction of methyl chloride (CH3Cl) with water to form methanol (CH3OH) and hydrochloric acid (HCl) is 3.32 x 10-10 s-1 at 25oC. Calculate the rate constant at 40oC if the activation energy is 116 kJ/mol.

23. Reaction Mechanisms and Rate Laws • The global reaction is often the sum of a series of simple reactions (elementary steps) • The sequence of elementary steps that give the product is the reaction mechanism • eg.; For the reaction 2 NO(g) + O2(g)  2 NO2(g) , the reaction mechanism is • The sum of the two elementary steps is the global reaction • In this example, N2O2 is an intermediate, i.e., a species that appears in the reaction mechanism but not in the overall balanced reaction • An intermediate is produced and then consumed Step 1: Step 2:

24. Reaction Mechanisms and Rate Laws • The number of molecules that react in an elementary step is the molecularity of that step • A unimolecular step involves a single molecule • A bimolecular step involves two molecules • A trimolecular step involves three molecules • A trimolecular step is rare because it is unlikely that three molecules will collide simultaneously • The rate law for an elementary step is given by the stoichiometry of the step, i.e., • For A  products, r = k[A] • For A + B  products, r = k[A][B] • For 2 A  products,r = k[A]2

25. Reaction Mechanisms and Rate Laws • A mechanism can be deduced from experimental observations • The reaction mechanism must obey two rules • The sum of the elementary steps must correspond to the balanced equation of the global reaction • The slowest elementary step of the series, the rate-determining step, should match that of the overall reaction • The rate-determining step determines the rate of the overall reaction

26. Reaction Mechanisms and Rate Laws • Take, for example, the reaction • All alone, this reaction is slow • However, with the presence of iodide ions, the reaction is fast and the rate law is • The mechanism cannot simply be two molecules of H2O2 that collide and then react

27. Reaction Mechanisms and Rate Laws • A possible mechanism is • If step 1 is the rate-determining step, the rate law for the global reaction would be • Such a rate law is observed experimentally, so the mechanism is plausible

28. Reaction Mechanisms and Rate Laws • In the proposed mechanism, IO- is an intermediate since it is produced in the first step and consumed in the second • In the proposed mechanism, I- is not an intermediate since it is present before the reaction occurs and is still present once the reaction is finished • I- is a catalyst

29. Reaction Mechanisms and Rate Laws • Example: We believe that the formation of NO and CO2 from NO2 and CO happens in two steps: The experimentally determined rate law is k[NO2]2. (a) Give the equation of the global reaction. (b) Indicate which species is an intermediate. (c) What can be said about the relative rates of Steps 1 and 2? • Solution: (a) The sum of Steps 1 and 2 is: NO2 + CO  NO + CO2 (b) NO3 is an intermediate (c) Step 1 is the slow step since its kinetics correspond to that of the global reaction (the value of [CO] would be in the rate law if Step 2 was the rate-determining step)

30. The reaction A(aq) B(aq) is a first-order reaction with respect to A. The half-life at 25.0oC is 875.3 s. If the activation energy of this reaction is 35.1 kJ/mol, what would the concentration of A be after 1000.0 s if we have an initial concentration of A of 0.500 M and temperature is held constant at 50.0oC?